1.10 g H 2 isallowed to react with 10.5 g N 2 , producing2.31 g NH 3 . 1-What is
ID: 686045 • Letter: 1
Question
1.10 g H2 isallowed to react with 10.5 g N2, producing2.31 g NH3. 1-What is the theoretical yield for thisreaction under the given conditions? 2-What is the percent yield for thisreaction under the given conditions? 1.10 g H2 isallowed to react with 10.5 g N2, producing2.31 g NH3. 1-What is the theoretical yield for thisreaction under the given conditions? 1-What is the theoretical yield for thisreaction under the given conditions? 2-What is the percent yield for thisreaction under the given conditions?Explanation / Answer
3H2 + N2 ------> 2NH3 Molar mass of NH3 is = 14 + 3 * 1= 17 g 1-What is the theoretical yield for this reaction under thegiven conditions? 3 * 2 g(=6) of H2 when reacted with 28g of N2 produces 2 *17(=34) g of NH3 6 g of H2 reacts with 28 g of N2 1.1 g of H2 reacts with X g ofN2 X = ( 1.1*28 ) / 6 = 5.1333 g of N2 So, 10.5 -5.1333 = 5.3667 g of N2 is left unreacted. 6 g of H2 produces 34 g of NH3 1.1 g of H2 produces Y g of NH3 Y = ( 1.1 * 34 ) / 6 =6.2333 g of NH3 So, the theoretical yield is 6.2333 g 2-What is the percent yield for this reaction under the givenconditions? % yield = ( actual yield / theoretical yield ) *100 = ( 2.31 / 6.2333) * 100 = 37.059 % ~ 37.1 % Molar mass of NH3 is = 14 + 3 * 1= 17 g 1-What is the theoretical yield for this reaction under thegiven conditions? 3 * 2 g(=6) of H2 when reacted with 28g of N2 produces 2 *17(=34) g of NH3 6 g of H2 reacts with 28 g of N2 1.1 g of H2 reacts with X g ofN2 X = ( 1.1*28 ) / 6 = 5.1333 g of N2 So, 10.5 -5.1333 = 5.3667 g of N2 is left unreacted. 6 g of H2 produces 34 g of NH3 1.1 g of H2 produces Y g of NH3 Y = ( 1.1 * 34 ) / 6 =6.2333 g of NH3 So, the theoretical yield is 6.2333 g 2-What is the percent yield for this reaction under the givenconditions? % yield = ( actual yield / theoretical yield ) *100 = ( 2.31 / 6.2333) * 100 = 37.059 % ~ 37.1 %Related Questions
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